UNIT – IV
SPACE AND CABLE STRUCTURES
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SPACE TRUSS
• A space frame or space truss is a three dimensional assemblage of
line members, each member being joined at its ends to the
foundation or to other members by frictionless ball and socket
joints. The simplest space frame consists of six members joined to
form a tetrahedron, Fig. By beginning with six members forming a
tetrahedron, a stable space frame can be constructed by successive
addition of three new members and a joint.
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• In Space Truss,
• Statically determinate if,
m + r = 3j
Statically Indeterminate if,
m + r > 3j
Where j is number of joints, m number of members, r is number of
restraints against translation
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Tension Coefficient Method
• The method is based on the fundamental equations of static
equilibrium and may be readily applied for the solution of
statically determinate pin jointed space frames
• It provides elegant and systematic procedure of formulating
the various simultaneous equation and especially suitable for
solving space trusses.
•
In this method the three equations of static equilibrium, one
along each of the three Cartesian coordinate axes, are written
at each joint.
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• If the applied forces , all but one member meeting at a joint lie
in one plane , the force in the member not in that plane will be
equal to zero.
• The number of base reaction components do not influence the
computation if the base joints are not interconnected.
• The computation can be proceed systematically from the end
joint with three connecting bars to other joints with not more
than three unknown member forces
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Consider a member AB, connecting joints A and B of a pin jointed space frame.
The length of member AB is given by the equation,
LAB = L= [(xB – xA) 2 + (yB – yA) 2 + (zB - zA) 2]1/2
Or
LAB = L= [(xA – xB) 2 + (yA – yB) 2 + (zA - zB) 2]1/2
The axial force in member AB may be expressed as
SAB = tAB LAB
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LAB = L= [(xB – xA) 2 + (yB – yA) 2 + (zB - zA) 2]1/2
Fx = F cos θx
cos θx = (XB - XA) / L
Fy = F cos θy
cos θY = (YB - YA) / L
Fz = F cos θz
cos θZ = (ZB - ZA) / L
Or
Fx = F (XB - XA) / L
Fy = F (YB - YA) / L
Fz = F (ZB - ZA) / L
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•
Suppose If there are more number of member meeting at a joint and with
external forces then Equilibrium Equations can be written as
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• F/L = t
If
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• Finally Member Force
Fi = ti Li
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Steps Involved
•1.
Write the co ordinates for all the members (x, y, z)
2.
Equate for every joints
3. Solve the above simultaneous equation and find Tension
coefficient ti
4.
Find the member forces Member Force Fi = ti Li
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Problem 1.
Using the tension coefficient method, calculate the forces in the members of
the pin jointed space frame shown in fig .The numbers in parentheses are the
Cartesian coordinates of the joints of the frame.
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• The lengths of members DA, DB, and DC may be computed
by using Equation,
•
•
•
LDA = [(0 - 12)2 + (0 - 6)2 + (0 - 8)2 ]1/2
LDB = [(8 - 12)2 + (5 - 6)2 + (0 - 8)2]1/2
LDC = [(0 - 12)2 + (10 - 6)2 + (0 - 8)2]1/2
= 15.620 m
= 9.000m
= 14.967 m
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• Components of the external forces acting at
joint D are,
XD = 40 kN
YD = 30kN
ZD = - 80 kN
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• Considering the equilibrium of joint D and using Equation,
•
tDA ( 0- 12) + tDB (8 - 12) + tDC (0 - 12) + 40 = 0
•
tDA ( 0- 6) + tDB (5 - 6) + tDC (10 - 6) + 30
=0
•
tDA ( 0- 8) + tDB (0 - 8) + tDC (0 - 18) - 80
=0
• Solving Eq. the tension coefficients,
• tDA = 9 kN/m tDB = - 20 kN/m tDC = 1 kN/m
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Member Forces
SDA = 140.58 kN
SDB = - 180kN (compressive)
SDC = 14.97 kN
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The external reaction components at the supports may be computed by
considering the equilibrium of joints A, B and C
•
•
•
•
9(12 - 0) + XA = 0
9(6 - 0) + YA = 0
9(8 - 0) + ZA = 0
XA = - 108 kN
YA = - 54kN
ZA = - 72 kN
The external reaction components at joints B and C can be calculated
in a similar manner.
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Problem 2
A space frame shown in Figure is supported at A, B, C and D in a horizontal
plane, through ball joints. The member EF is horizontal, and is at a height of 3
m above the base. The loads at the joints E and F, shown in the figure act in a
horizontal plane. Find the forces in all the members of the frame.
Point
X
Y
Z
B
0
0
0
C
7
0
0
D
7
6
0
A
0
6
0
E
2
3
3
F
5
3
3
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• LAE = [(2 - 0)2 + (3 - 6)2 + (3 - 0)2]1/2
= 4.6904 m = LBE = LCF = LDF
• LBF = [(5 - 0)2 + (3 - 0)2 + (3 - 0)2 ]1/2
= 6.5574 m
• LEF = 3 m (given)
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• Joint E:
The three equations at joint E as follows
• tEA (xA – xE) + tEB (xB - xE) + tEF (xF - xE) + XE = 0
• tEA (yA – yE) + tEB (yB - yE) + tEF (yF - yE) + YE = 0
• tEA (zA – zE) + tEB (zB - zE) + tEF (zF - zE) + ZE = 0
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• -2 tEA – 2 tEB + 3 tEF + 5 = 0
(1)
• -3 tEA – 3 tEB + 10 = 0
(2)
• -3 tEA – 3 tEB + = 0
(3)
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• From (3),
tEA = - tEB
• From (2),
tEA + tEA = - 10/3
tEA = - 5/3
tEB = + 5/3
Or
• From (1),
•
+ 3 tEF + 5 = 0
tEF = - 5/3
(a)
(b)
(c)
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Joint F:
The three equations are
•
•
•
tFD (xD – xF) + tFC (xC – xF) + tFB (xB – xF) + tFE (xE – xF) + XF = 0
tFD (yD – yF) + tFC (yC – yF) + tFB (yB – yF) + tFE (yE – yF) + YF = 0
tFD (zD – zF) + tFC (zC – zF) + tFB (zB – zF) + tFE (zE – zF) + ZF = 0
• 2 tFD + 2 tFC -5 tFB – 3 tFE = 0
• 3 tFD - 3 tFC -3 tFB + 15 = 0
• -3tFD - 3 tFC -3 tFB = 0
tFD = -2.5
• tFB = + 0.7143
• tFC = + 1.7857
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Tension Coefficients and Forces in the various Members
Member
Length
t
T (kN)
EA
4.6904
- 1.667
- 7.817
EB
4.6904
+1.6667
+7.817
EF
3.0
- 1.6667
+ 5.0
FC
4.6904
+1.7857
+ 8.376
FD
4.6904
- 2.50
- 11.726
FB
6.5574
+0.7143
+ 4.684
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Displacement of Space Truss
Ss L
 
AE
j
j
• Where L/AE is flexibility of member
S is member forces due to external load
Sj is member forces due to dummy unit load
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THANK YOU
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